Arrange the given terms to create a true exponential equation and a true logarithmic equation.

Transcription

1 Lesson.1 Skills Practice Name Date All the Pieces of the Puzzle Exponential and Logarithmic Forms Vocabulary 1. Give an example of a logarithmic equation. Answers will vary. Sample answer: log Problem Set Arrange the given terms to create a true exponential equation and a true logarithmic equation , 5, , 11, 31 Exponential equation: Exponential equation: Logarithmic equation: log Logarithmic equation: log , 256, , 4, 6 Exponential equation: Exponential equation: Logarithmic equation: log Logarithmic equation: log , 1 512, 8 Exponential equation: Logarithmic equation: log 8 ( ) , 4, 16 2 Exponential equation: Logarithmic equation: log , 1, Exponential equation: Logarithmic equation: log 17 ( ) , 9, 729 Exponential equation: Logarithmic equation: log Chapter Skills Practice 735

2 Lesson.1 Skills Practice page 2 Solve for the unknown. 9. log n log n 7 n n n log n 4 log n 4 ( 1 4 ) n 5 64 ( 1 4 ) n 5 43 ( 1 4 ) n 5 ( 1 4 ) 23 n log n log n n n n log n ( ) 5 24 log n ( ) 5 24 n n 24 5 ( 1 5 ) 4 n n 5 5. log n ( 4 8 ) log n ( 4 8 )5 4 n n n (2 3 ) 1 4 n log n 5 6 log n n 1,000,000 5 n n log n log n 4 n n n log 81 ( 1 9 ) 5 n log 81 ( 1 9 ) 5 n 1 81 n 5 9 (9 2 ) n n n 5 21 n Chapter Skills Practice

3 Lesson.1 Skills Practice page 3 Name Date Estimate the logarithm to the tenths place. Explain how you determined your answer. 17. log 2 48 The logarithm log 2 48 is approximately equal to 5.5. log 2 32, log 2 48, log , n, 6 Because 48 is halfway between 32 and 64, the approximation should be halfway between 5 and log 3 2 Answers will vary. The logarithm log 3 2 is approximately equal to 0.5. log 3 1, log 3 2, log 3 3 0, n, 1 Because 2 is halfway between 1 and 3, the approximation should be halfway between 0 and log 8 ( ) Answers will vary. The logarithm log 8 ( ) is approximately equal to log 8 ( ), log 8 ( ), log 8 ( 1 64 ) 23, n, 22 Because 1 is closer to 1 than 1, the approximation should be closer to 23 than log 6 53 Answers will vary. The logarithm log 6 53 is approximately equal to 2.1. log 6 36, log 6 53, log , n, 3 Because 53 is closer to 36 than 216, the approximation should be closer to 2 than 3. Chapter Skills Practice 737

4 Lesson.1 Skills Practice page log Answers will vary. The logarithm log is approximately equal to 3.9. log 9 729, log , log , n, 4 Because 6000 is closer to 6561 than 729, the approximation should be closer to 4 than log 7 ( 1 10 ) Answers will vary. The logarithm log 7 ( 1 10 ) is approximately equal to log 7 ( 1 7 ) 49 ), log 7 ( 1 10 ), log 7 ( 1 22, n, 21 Because 1 is closer to 1 than 1, the approximation should be closer to 21 than log 1 ( ) Answers will vary. The logarithm log 1 ( ) is approximately equal to 4.5. log 1 ( ), log1 ( ), log1 ( ) 4, n, 5 Because 1 is halfway between 1 and 1, the approximation should be halfway between and log Answers will vary. The logarithm log 4 is approximately equal to log , log4 0.85, log , n, 0 Because 0.85 is closer to 0.75 than 1, the approximation should be closer to 21 than Chapter Skills Practice

5 Lesson.1 Skills Practice page 5 Name Date Determine the appropriate base of each logarithm. Explain your reasoning. 25. log b log 5 25, log b 29, log , 2.1, 3 Because the value of the exponent is 2.1, that means that the argument 29 should be close to its lower limit. When the base is 5, 29 is very close to the lower limit of 25, whereas when the base is 4, 29 is not very close to the lower limit of log b log 6 6, log b 35, log , 1.9, 2 Because the value of the exponent is 1.9, that means that the argument 35 should be close to its upper limit. When the base is 6, 35 is very close to the upper limit of 36, whereas when the base is 7, 35 is not very close to the upper limit of log b ( 1 7 ) log 4 ( 1 8 ), log b ( 1 7 ), log 4 ( 1 4 ) 22, 21.9, 21 1 Because the value of the exponent is 21.9, that means that the argument should be close to its 7 lower 1 1 limit. When the base is 4, is very close to the lower limit of 1, whereas when the base 7 8 is 5, is not very close to the lower limit of Chapter Skills Practice 739

6 Lesson.1 Skills Practice page log b log 3 27, log b 80, log , 3.9, 4 Because the value of the exponent is 3.9, that means that the argument 80 should be close to its upper limit. When the base is 3, 80 is very close to the upper limit of 81, whereas when the base is 4, 80 is not very close to the upper limit of log b log 7 1, log b 6, log 7 7 0, 0.9, 1 Because the value of the exponent is 0.9, that means that the argument 6 should be close to its upper limit. When the base is 7, 6 is very close to the upper limit of 7, whereas when the base is 8, 6 is not very close to the upper limit of log b log 4 64, log b 66, log , 3.1, 4 Because the value of the exponent is 3.1, that means that the argument 66 should be close to its lower limit. When the base is 4, 66 is very close to the lower limit of 64, whereas when the base is 3, 66 is not very close to the lower limit of Chapter Skills Practice

7 Lesson.2 Skills Practice Name Date Mad Props Properties of Logarithms Problem Set Rewrite each logarithmic expression in expanded form using the properties of logarithms. 1. log 3 ( 5x ) log 3 (5x) 5 log log 3 x 2. log 5 ( a b ) log 5 ( a b ) 5 log 5 a 2 log 5 b 3. log 7 ( n 4 ) log 7 ( n 4 )5 4 log 7 n 4. log ( x 7 ) log ( x 7 ) 5 log x 2 log 7 5. log 2 ( mn ) 6. log ( p q ) log 2 (mn) 5 log 2 m 1 log 2 n log (p q ) 5 q log p 7. ln ( x 2 ) In (x 2 ) 5 2 ln x 8. ln ( c 3 ) ln ( c 3 ) 5 ln c 2 ln 3 9. log 3 ( 7x 2 ) 10. ln ( 2x 3 y 2 ) log 3 ( 7x 2 )5 log log 3 x ln ( 2x 3 y 2 )5 ln ln x 1 2 ln y 11. log ( xy 5 ) log ( xy 5 ) 5 log x 1 log y 2 log log 7 ( 3x4 y ) log 7 ( 3x4 y ) 5 log3 1 4 logx 2 logy ln ( x 7y ) ln ( x 7y ) 5 ln x 2 ln 7 2 ln y 14. log 5 ( 7x2 y 3 ) log 5 ( 7x2 y 3 ) 5 log7 1 2 logx 2 3 logy log ( xyz ) log (xyz) 5 log x 1 log y 1 log z 16. ln ( x 1 1 _ (y 1 3) 2 ) _ ln ( x 1 1 (y 1 3) 2 ) 5 ln (x 1 1) 2 2 ln (y 1 3) Chapter Skills Practice 741

8 Lesson.2 Skills Practice page 2 Rewrite each logarithmic expression as a single logarithm. 17. log x 2 2 log y log x 2 2 log y 5 log ( x y 2 ) log 4 x 1 log 4 y 2 log 4 z 3 log 4 x 1 log 4 y 2 log 4 z 5 log 4 ( x3 y z ) log 2 x 2 2 log 2 x 6 log 2 x 2 2 log 2 x 5 log 2 ( x6 x 2 ) 5 log ( x 4 ) log log 7 2 log 6 log log 7 2 log 6 5 log ( 3(7)2 6 ) 5 log ( ) 5 log log x 1 3 log y log z log x 1 3 log y 2 1 ) 2 ( log z 5 log xy3 z log 3 x 2 (2 log 3 x 1 5 log 3 y) 7 log 3 x 2 (2 log 3 x 1 5 log 3 y) 5 log 3 ( x7 y 5 ) x 2 y 5 ) 5 log 3 ( x ln (2x 1 3) 2 4 ln (y 2 2) 2 ln (2x 1 3) 2 4 ln (y 2 2) 5 ln ( (2x 1 3)2 ( y 2 2) 4 ) 24. ln (x 2 7) 2 2(ln x 1 ln y) ln (x 2 7) 2 2(ln x 1 ln y) 5 ln ( x 2 7 x 2 y 2 ) Suppose w 5 log b 2, x 5 log b 3, y 5 log b 7, and z 5 log b 11. Write an algebraic expression for each logarithmic expression. 25. log b log b 98 log b 33 5 log b (3? 11) log b 98 5 log b (7 2? 2) 5 log b 3 1 log b log b 7 1 log b 2 5 x 1 z 5 2y 1 w 27. log b ( 2 3 ) log b ( 2 3 ) 5 log b 2 2 log b 3 5 w 2 x 28. log b ( 7 8 ) log b ( 7 8 ) 5 log b 7 2 log b (23 ) 5 log b log b 2 5 y 2 3w 29. log b 1.5 log b log b ( 3 2 ) 5 log b 3 2 log b log b 2.75 log b log b ( ) 5 log b log b 2 5 x 2 w 5 z 2 2w 742 Chapter Skills Practice

9 Lesson.3 Skills Practice Name Date What s Your Strategy? Solving Exponential Equations Vocabulary 1. Define the Change of Base Formula and explain how it is used. The Change of Base Formula states: log b c 5 logc a, where a, b, c. 0 and a, b fi 1. It allows you log a b to calculate an exact value for a logarithm by rewriting it in terms of a different base. Problem Set Solve each exponential equation by using the Change of Base Formula x x x x x log 6 24 log 24 x log 6 x log log 65 x log 12 x x x x x x x x x 5 log 7 15 log 15 3x 5 log 7 3x x x 5 log 8 71 log 71 5x 5 log 8 5x x 0.41 Chapter Skills Practice 743

10 Lesson.3 Skills Practice page x x x x x x x log 4 39 log 39 x log 4 x x log log 51 x log 11 x x x ( 2 3 ) 3x ( 2 3 ) 3x ( 2 3 ) 3x ( 3 5 ) 2x ( 3 5 ) 2x ( 3 5 ) 2x x 5 log log 62 3x 5 log ( 2 3 ) 3x x 5 log _ log 111 2x 5 log ( 3 5 ) 2x x x (5) 2x (2) x (5) 2x (2) x (5) 2x (2) x (5) 2x x log 5 7 2x log 7 log 5 2x (2) x x log 2 9 x log 9 log 2 x x x x Chapter Skills Practice

11 Lesson.3 Skills Practice page 3 Name Date Solve each exponential equation using properties of logarithms x x x x (x 2 2) log 7 5 log 28 log 28 x log 7 (x 1 2) log 15 5 log 60 log 60 x log 15 x x x x x x x x x log 5 5 log 32 log 32 4x 5 log 5 3x log 9 5 log 124 _ log 124 3x 5 log 9 4x x x x x x x x (x 1 7) log 3 5 log 68 log 68 x log 3 14 x x (x 2 6) log 14 5 log 70 log 70 x log 14 x x x x Chapter Skills Practice 745

12 Lesson.3 Skills Practice page ( 4 7 ) 2x ( 4 7 ) 2x ( 4 7 ) 2x x log ( 4 7 ) 5 log 58 log ( 4 7 ) log 58 2x 5 2x x ( 5 8 ) 4x ( 5 8 ) 4x ( 5 8 ) 4x x log ( 5 8 ) 5 log 56 log ( 5 8 ) log 56 4x 5 4x x (4) 3x (4) 3x (4) 3x (4) 3x (3x 2 6) log 4 5 log 11 log 11 3x log 4 3x x x (3) x (3) x (3) x (3) x (x 1 6) log 3 5 log 22 log 22 x log 3 x x Chapter Skills Practice

13 Lesson.3 Skills Practice page 5 Name Date Solve each exponential equation. Explain why you chose the method that you used x x (x 2 4) log 11 5 log 343 _ log 343 x log 11 x x I took the log of both sides, because 343 cannot be written as a power of x Answers will vary. 5 2x x x x 5 4 x 5 2 I used common bases, because 3125 can be written as ( 8 9 ) 5x Answers will vary. 16 ( 8 9 ) 5x ( 8 9 ) 5x x log ( 8 9 ) 5 log 47 log 47 5x 5 log ( 8 9 ) 5x x I took the log of both sides, because 752 cannot be written as a power of 8 9. Chapter Skills Practice 747

14 Lesson.3 Skills Practice page x Answers will vary. 23 5x x 5 log _ log 736 5x 5 log 23 5x x I used the Change of Base Formula, because 736 cannot be written as a power of ( 7 12 ) 4x 5 49 Answers will vary. ( 7 12 ) 4x x 5 log log 49 4x 5 log ( 7 12 ) 4x x I used the Change of Base Formula, because 49 cannot be written as a power of x Answers will vary. 14 5x x x x x 5 15 x 5 3 I used common bases, because 2744 can be written as Chapter Skills Practice

15 Lesson.4 Skills Practice Name Date Logging On Solving Logarithmic Equations Problem Set Solve each logarithmic equation. Check your answer(s). 1. log 2 (x 2 2 x) 5 1 log 2 (x 2 2 x) x 2 2 x 0 5 x 2 2 x (x 1 1)(x 2 2) x 5 21, 2 Check: log 2 ( (21) 2 2 (21) )0 1 log 2 (1 1 1) 0 1 log log 2 ( ) 0 1 log 2 (4 2 2) 0 1 log log 15 (x 2 2 2x) 5 1 log 15 (x 2 2 2x) x 2 2 2x 0 5 x 2 2 2x (x 1 3)(x 2 5) x 5 23, 5 Check: log 15 ((23) 2 2 2(23)) 0 1 log 15 (9 1 6) 0 1 log log 15 ( (5)) 0 1 log 15 ( ) 0 1 log Chapter Skills Practice 749

16 Lesson.4 Skills Practice page 2 3. log 6 (x 2 1 5x) log 2 (x 2 1 6x) 5 4 log 6 (x 2 1 5x) 5 2 log 2 (x 2 1 6x) x 2 1 5x x 2 1 6x 36 5 x 2 1 5x 16 5 x 2 1 6x 0 5 x 2 1 5x x 2 1 6x (x 1 9)(x 2 4) 0 5 (x 1 8)(x 2 2) x 5 29, 4 x 5 28, 2 Check: Check: log 6 ((29) 2 1 5(29)) 0 2 log 2 ((28) 2 1 6(28)) 0 4 log 6 ( ) 0 2 log 2 ( ) 0 4 log log log 6 ( (4)) 0 2 log 2 ( (2)) 0 4 log 6 ( ) 0 2 log 2 (4 1 12) 0 4 log log log 4 (x x) log 10 (x x) 5 2 log 4 (x x) 5 3 log 10 (x x) x x x x 64 5 x x x x 0 5 x x x x (x 1 4)(x 2 16) 0 5 (x 1 20)(x 2 5) x 5 24, 16 x 5 220, 5 Check: Check: log 4 ((24) (24)) 0 3 log 4 ( ) 0 3 log log 4 ( (16)) 0 3 log 10 ((220) (220)) 0 2 log 10 ( ) 0 2 log log 10 ( (5)) 0 2 log 4 ( ) 0 3 log 10 ( ) 0 2 log log Chapter Skills Practice

17 Lesson.4 Skills Practice page 3 Name Date 7. log 3 (3x 2 118x) 5 4 log 3 (3x x) x x x x 0 5 3x x (x 2 1 6x 2 27) 0 5 3(x 1 9)(x 2 3) x 5 29, 3 Check: log 3 (3(29) (29)) 0 4 log 3 ( ) 0 4 log log 3 (3(3) (3)) 0 4 log 3 ( ) 0 4 log log 4 (2x x) 5 3 log 4 (2x x) x x x x 0 5 2x x (x x 2 32) 0 5 2(x 1 2)(x 2 16) x 5 22, 16 Check: log 4 (2(22) (22)) 0 3 log 4 (8 1 56) 0 3 log log 4 (2(16) (16)) 0 3 log 4 ( ) 0 3 log Chapter Skills Practice 751

18 Lesson.4 Skills Practice page 4 Use the properties of logarithms to solve each logarithmic equation. Check your answer(s) log 3 x 2 log log 3 (x 2 2) 2 log 3 x 2 log log 3 (x 2 2) log 3 x 2 2 log log 3 (x 2 2) log 3 ( x2 8 ) 5 log (x 2 2) 3 x 2 2 8x Check: x2 8 5 x 2 2 x 2 5 8x 2 16 (x 2 4) x log log log 3 (4 2 2) log log log 3 2 log 3 ( 16 8 ) 0 log2 3 log log log 4 (x 1 3) 1 log 4 x 5 1 log 4 (x 1 3) 1 log 4 x 5 1 log 4 (x(x 1 3)) 5 1 x(x 1 3) x 2 1 3x 5 4 x 2 1 3x (x 1 4)(x 2 1) 5 0 x 5 24, 1 Check: 24 is an extraneous solution. log 4 (1 1 3) 1 log log log Chapter Skills Practice

19 Lesson.4 Skills Practice page 5 Name Date 11. log (2x 2 1 3) 1 log 2 5 log 10x log (2x 2 1 3) 1 log 2 5 log 10x log (2(2x 2 1 3)) 5 log 10x 2(2x 2 1 3) 5 10x 2x x 2x 2 2 5x (2x 2 3)(x 2 1) 5 0 x 5 3 2, 1 Check: log 2 ( 3 2 ) log 2 0 log 10 ( 3 log ( 15 2 ) 1 log 2 0 log 15 log ( ) 0 log 15 log 15 5 log 15 2 ) 12. log 2 x 1 log 2 (x 2 6) 5 4 log 2 x 1 log 2 (x 2 6) 5 4 log 2 (x(x 2 6)) 5 4 x(x 2 6) x 2 2 6x 5 16 x 2 2 6x (x 2 8)(x 1 2) 5 0 x 5 8, 22 Check: 22 is an extraneous solution. log log 2 (8 2 6) 0 4 log log log 2(1) log 2 0 log (10(1)) log 5 1 log 2 0 log 10 log ((5)2) 0 log 10 log 10 5 log 10 Chapter Skills Practice 753

20 Lesson.4 Skills Practice page 6. 2 log 5 x 2 log log 5 (8 2 x) 2 log 5 x 2 log log 5 (8 2 x) log 5 (x 2 ) 2 log log 5 (8 2 x) log 5 ( x2 4 ) 5 log (8 2 x) 5 x 2 1 4x (x 1 8)(x 2 4) 5 0 Check: x x x x x 5 28, 4 28 is an extraneous solution. 2 log log log 5 (8 2 4) log 5 ( 4 2 )2 log log 5 4 log log log 5 4 log 5 ( 16 4 ) 0 log4 5 log log log log 2 (3x 2 1 4) 5 log 2 ( 39x ) log log 2 (3x 2 1 4) 5 log 2 ( 39x ) Check: log 2 (3(3x 2 1 4)) 5 log 2 ( 39x ) 3(3x 2 1 4) 5 39x 3x x 3x 2 2 x (3x 2 1)(x 2 4) 5 0 x 5 1 3, 4 log log 2 ( 3 ( 1 3 ) ) 0 log 2 39 ( 1 log 2 ( 3 ( 3 ( 1 9 ) 1 4 ) ) 0 log 2 log 2 ( 3 ( ) ) 0 log 2 log 2 (1 1 12) 0 log 2 log 2 5 log 2 log log 2 (3(4) 2 1 4) 0 log 2 39(4) log 2 (3(3(16) 1 4)) 0 log log 2 (3(48 1 4)) 0 log log 2 (3(52)) 0 log log log ) 754 Chapter Skills Practice

21 Lesson.4 Skills Practice page 7 Name Date 15. ln ( x ) 1 ln 2 5 ln (11x) ln ( x ) 1 ln 2 5 ln (11x) ln ( 2 ( x ) ) 5 ln (11x) ln (2x ) 5 ln (11x) 2x x 2x x (2x 2 5)(x 2 3) 5 0 x 5 5 2, log 4 ( 1 5 x2 2 6 ) 2 log 4 ( 1 5 ) 5 logx 4 log 4 ( 1 5 x2 2 6 ) 2 log 4 ( 1 5 ) 5 logx 4 1 log 4 5 ( x )5 log 4 x log 4 ( 5 ( 1 5 x2 2 6 ) ) 5 log 4 x log 4 (x ) 5 log 4 x x x x 2 2 x Check: ln ( ( 5 2 ) ) 1 ln 2 0 ln ( 11 ( 5 2 ) ) ln ( 2 ( ) ) 0 ln ( 55 2 ) ln ( ) 0 ln ( 55 2 ) ln ( 55 2 ) 5 ln ( 55 2 ) ln ( ) 1 ln 2 0 ln (11(3)) ln ( 2 ( ) ) 0 ln 33 ln ( ) 0 ln 33 ln 33 5 ln 33 Check: (x 1 5)(x 2 6) 5 0 x 5 25, 6 25 is an extraneous solution. log 4 ( 1 5 (6)2 2 6 ) 2 log 4 ( 1 5 ) 0 log6 4 log 4 ( 1 5 (36) 2 6 ) 2 log 4 ( 1 5 ) 0 log6 4 log 4 ( ) 2 log 4 ( 1 5 ) 0 log6 4 log 4 ( ) 2 log 4 ( 1 5 ) 0 log6 4 log 4 ( 6 5 ) 2 log 4 ( 1 5 ) 0 log6 4 log 4 ( )0 log log log 4 6 Chapter Skills Practice 755

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23 Lesson.5 Skills Practice Name Date So When Will I Use This? Applications of Exponential and Logarithmic Equations Problem Set The amount of a radioactive isotope remaining can be modeled using the formula A 5 A 0 e 2kt, where t represents the time in years, A represents the amount of the isotope remaining in grams after t years, A 0 represents the original amount of the isotope in grams, and k is the decay constant. Use this formula to solve each problem. 1. Strontium-90 is a radioactive isotope with a half-life of about 29 years. Calculate the decay constant for Strontium-90. Then find the amount of 100 grams of Strontium-90 remaining after 120 years. A 5 A 0 e 2kt 1 2 A 1 5 A 0 0 e2k(29) 2 5 e229k ln ( k 2 ) k The decay constant for Strontium-90 is about A 5 100e (120) A After 120 years, there would be about grams remaining. 2. Radium-226 is a radioactive isotope with a half-life of about 1622 years. Calculate the decay constant for Radium-226. Then find the amount of 20 grams of Radium-226 remaining after 500 years. A 5 A 0 e 2kt 1 2 A 1 5 A 0 0 e2k(1622) 2 5 e21622k ln ( k 2 ) k The decay constant for Radium-226 is about A 5 20e (500) A After 500 years, there would be about grams remaining. Chapter Skills Practice 757

24 Lesson.5 Skills Practice page 2 3. Carbon-14 is a radioactive isotope with a half-life of about 5730 years. Calculate the decay constant for Carbon-14. Then find the amount of 6 grams of Carbon-14 that will remain after 22,000 years. A 5 A 0 e 2kt 1 2 A 0 5 A 0 e 2k(5730) e25730k ln ( k 2 ) k The decay constant for Carbon-14 is about A 5 6e (22,000) A After 22,000 years, there would be about gram remaining. 4. Cesium-7 is a radioactive isotope with a half-life of about 30 years. Calculate the decay constant for Cesium-7. Then calculate the percentage of a Cesium-7 sample remaining after 100 years. A 5 A 0 e 2kt 1 2 A 1 5 A 0 0 e2k(30) 2 5 e230k ln ( k 2 ) k The decay constant for Cesium-7 is about A 5 A 0 e t A A 5 A 0 e (100) A 0 After 100 years, there will be about 9.93% of the Cesium-7 remaining. 758 Chapter Skills Practice

25 Lesson.5 Skills Practice page 3 Name Date 5. Uranium-232 is a radioactive isotope with a half-life of about 69 years. Calculate the decay constant for Uranium-232. Then calculate the percentage of a Uranium-232 sample remaining after 200 years. A 5 A 0 e 2kt 1 2 A 1 5 A 0 0 e2k(69) 2 5 e269k ln ( k 2 ) k The decay constant for Uranium-232 is about A 5 A 0 e t A A 5 A 0 e (200) 0.5 A 0 After 200 years, there will be about.5% of the Uranium-232 remaining. 6. Rubidium-87 is a radioactive isotope with a half-life of about years. Calculate the decay constant for Rubidium-87. Then calculate the percentage of a Rubidium-87 sample remaining after 1,000,000 years. A 5 A 0 e 2kt 1 2 A5 A ) e2k( )k e( ln ( 1 2 ) 5 ( )k k The decay constant for Rubidium-87 is about A 5 A 0 e 2( )t A A 5 A 0 e 2( (1,000,000) A 0 After 1,000,000 years, there will be about 98.5% of the Rubidium-87 remaining. Chapter Skills Practice 759

26 Lesson.5 Skills Practice page 4 Use the given exponential equation to answer each question. Show your work. 7. The number of students exposed to the measles at a school can be modeled by the equation S 5 10e 0.15t, where S represents the number of students exposed after t days. How many students were exposed after eight days? S 5 10e 0.15t 5 10e (0.15? 8) 5 10e Approximately 33 students were exposed after eight days. 8. The minnow population in White Mountain Lake each year can be modeled by the equation M 5 700(10 0.2t ), where M represents the minnow population t years from now. What will the minnow population be in 15 years? M 5 700(10 0.2t ) 5 700(10 (0.2?15) ) 5 700(10 3 ) 5 700(1000) 5 700,000 There will be 700,000 minnows in 15 years. 9. Aiden invested $600 in a savings account with continuous compound interest. The equation V 5 600e 0.05t can be used to predict the value, V, of Aiden s account after t years. What would the value of Aiden s account be after five years? V 5 600e 0.05t 5 600e (0.05? 5) 5 600e The value of Aiden s account would be about $ after five years. 760 Chapter Skills Practice

27 Lesson.5 Skills Practice page 5 Name Date 10. The rabbit population on Hare Island can be modeled by the equation R 5 60e 0.09t, where R represents the rabbit population t years from now. How many years from now will the rabbit population of Hare Island be 177 rabbits? R 5 60e 0.09t e 0.09t e 0.09t ln ln e 0.09t ln t ln e _ ln t ln e t t In a little longer than 12 years from now, there will be 177 rabbits on Hare Island. 11. A disease is destroying the elm tree population in the Dutch Forest. The equation N 5 16( t ) can be used to predict the number of elm trees, N, killed by the disease t years from now. In how many years from now will 406 elm trees have been killed by the disease? N 5 16( t ) ( t ) t log log t log t log 10 log t (1) t t In approximately 9.4 years, 406 elm trees will have been killed by the disease. Chapter Skills Practice 761

28 Lesson.5 Skills Practice page Manuel invested money in a savings account with continuous compound interest. The equation V 5 10,000e 0.03t can be used to determine the value, V, of the account after t years. In how many years will the value of the account be $12,000? V 5 10,000e 0.03t 12, ,000e 0.03t e 0.03t ln ln e 0.03t ln t ln e ln t(1) t t In approximately 6.1 years, the account will have a value of $12,000. Use the formula M 5 log ( I I 0 ), where M is the magnitude of an earthquake on the Richter scale, I 0 represents the intensity of a zero-level earthquake the same distance from the epicenter, and I is the number of times more intense an earthquake is than a zero-level earthquake, to solve each problem. A zero-level earthquake has a seismographic reading of millimeter at a distance of 100 kilometers from the center.. An earthquake southwest of Chattanooga, Tennessee in 2003 had a seismographic reading of millimeters registered 100 kilometers from the center. What was the magnitude of the Tennessee earthquake of 2003 on the Richter scale? M 5 log ( I I 0 ) M 5 log ( ) M 4.9 The Tennessee earthquake of 2003 measured 4.9 on the Richter scale. 14. An earthquake in Illinois in 2008 had a seismographic reading of millimeters registered 100 kilometers from the center. What was the magnitude of the Illinois earthquake of 2008 on the Richter scale? M 5 log ( I I 0 ) M 5 log ( ) M 5.2 The Illinois earthquake of 2008 measured 5.2 on the Richter scale. 762 Chapter Skills Practice

29 Lesson.5 Skills Practice page 7 Name Date 15. An earthquake off the northern coast of California in 2005 had a seismographic reading of 15,849 millimeters registered 100 kilometers from the center. What was the magnitude of the California earthquake in 2005 on the Richter scale? M 5 log ( I I 0 ) _ M 5 log ( 15, ) M 7.2 The California earthquake of 2005 measured 7.2 on the Richter scale. 16. The devastating earthquake in Haiti in 2010 had a magnitude of 7.0 on the Richter scale. What was its seismographic reading in millimeters 100 kilometers from the center? M 5 log ( I I 0 ) I log ( I ) I 10,000 The seismographic reading was about 10,000 millimeters. 17. Calculate the value of the seismographic reading for an earthquake of magnitude 6.4 on the Richter scale. M 5 log ( I I 0 ) I log ( I ) I 2512 The seismographic reading is about 2512 millimeters. Chapter Skills Practice 763

30 Lesson.5 Skills Practice page Calculate the value of the seismographic reading for an earthquake of magnitude 8.1 on the Richter scale. M 5 log ( I I 0 ) I log ( I ) I 125,893 The seismographic reading was about 125,893 millimeters. Use the given formula to solve each problem. 19. The formula for the population of a species is n 5 k log (A), where n represents the population of a species, A is the area of the region in which the species lives, and k is a constant that is determined by field studies. Based on population samples, an area that is 1000 square miles contains 360 wolves. Calculate the value of k. Then use the formula to find the number of wolves remaining in 15 years if only 300 square miles of this area is still inhabitable. n 5 k log (A) k log 1000 k The value of k is 120. n log (A) log In 15 years, there will be approximately 297 wolves remaining in the area. 20. The formula for the population of a species is n 5 k log (A), where n represents the population of a species, A is the area of the region in which the species lives, and k is a constant that is determined by field studies. Based on population samples, a rainforest that is 100 square miles contains 342 monkeys. Calculate the value of k. Then use the formula to find the number of monkeys remaining in 5 years if only 40 square miles of the rainforest survives due to the current level of deforestation. n 5 k log (A) k log 100 k The value of k is 171. n log (A) n log 40 n 274 In 5 years, there will be approximately 274 monkeys remaining in the area. 764 Chapter Skills Practice

31 Lesson.5 Skills Practice page 9 Name Date 21. The formula y 5 a 1 b ln t, where t represents the time in hours, y represents the amount of fresh water produced in t hours, a represents the amount of fresh water produced in one hour, and b is the rate of production, models the amount of fresh water produced from salt water during a desalinization process. In one desalination plant, cubic yards of fresh water can be produced in one hour with a rate of production of How much fresh water can be produced after 8 hours? y 5 a 1 b ln t y ln 8 y About cubic yards of fresh water can be produced in eight hours. 22. The formula y 5 a 1 b ln t, where t represents the time in hours, y represents the amount of fresh water produced in t hours, a represents the amount of fresh water produced in one hour, and b is the rate of production, models the amount of fresh water produced from salt water during a desalinization process. At a desalination plant, cubic yards of fresh water can be produced in one hour with a rate of production of How long will it take for the plant to produce 250 cubic yards of fresh water? y 5 a 1 b ln t ln t ln t t 817 It will take about 817 hours to produce 250 cubic yards of fresh water. Chapter Skills Practice 765

32 Lesson.5 Skills Practice page The relationship between the age of an item in years and its value is given by the equation log t 5 ( V C ), where t represents the age of the item in years, V represents the value of the item log (1 2 r) after t years, C represents the original value of the item, and r represents the yearly rate of appreciation expressed as a decimal. A luxury car was originally purchased for $110,250 and is currently valued at $65,200. The average rate of depreciate for this car is 10.3% per year. How old is the car to the nearest tenth of a year? log ( V C ) t 5 log (1 2 r) log ( 65, ,250 ) 5 log ( ) log log The car is approximately 4.8 years old. 24. The relationship between the age of an item in years and its value is given by the equation log t 5 ( V C ), where t represents the age of the item in years, V represents the value of the item after log (1 2 r) t years, C represents the original value of the item, and r represents the yearly rate of appreciation expressed as a decimal. A 4-year old car was originally purchased for $35,210. Its current value is $16,394. What is this car s annual rate of depreciation to the nearest tenth? log t 5 ( V C ) log (1 2 r) log ( 16,394 35,210 ) 4 5 log (1 2 r) 4 log (1 2 r) log log (1 2 r) r r r r This car s annual rate of depreciation is approximately 17.4%. 766 Chapter Skills Practice